Question

For one mole of an ideal gas an isochore is obtained. The slope of the isochore is \(0.082 \, \text{atm K}^{-1}\). What will be its pressure (in atm) when the temperature is 12.2 K? (R = \(0.082 \, \text{L atm mol}^{-1} \text{ K}^{-1}\)).

• A. 10.0
• B. 0.1
• C. 1.0
• D. 0.5

Hint:

- Isochoric process means constant volume. - Ideal Gas Law: \( PV=nRT \). - For an isochoric process, \( P = (\frac{nR}{V})T \). If P is plotted against T (in Kelvin), the graph is a straight line through the origin with slope \( \frac{nR}{V} \). - Given the slope and other parameters, you can find the constant volume V. - Then use \(PV=nRT\) or \(P = (\text{slope}) \times T\) to find pressure at a specific temperature.

Answer 1

The Correct Option is C

An isochore represents a process at constant volume (isochoric process).
For an ideal gas, \( PV = nRT \).
Since the volume V is constant and \(n=1\) mole, we can write \( P = \left(\frac{nR}{V}\right)T \).
This shows that P is directly proportional to T for an isochoric process.
The graph of P vs T is a straight line passing through the origin (if T is in Kelvin).
The slope of this line (isochore on a P-T diagram) is \( \frac{P}{T} = \frac{nR}{V} \).
Given slope \( = 0.
082 \, \text{atm K}^{-1} \).
So, \( \frac{nR}{V} = 0.
082 \, \text{atm K}^{-1} \).
We are given \( n=1 \) mole and \( R = 0.
082 \, \text{L atm mol}^{-1} \text{ K}^{-1} \).
\[ \frac{(1 \, \text{mol}) \times (0.
082 \, \text{L atm mol}^{-1} \text{ K}^{-1})}{V} = 0.
082 \, \text{atm K}^{-1} \] \[ \frac{0.
082 \, \text{L atm K}^{-1}}{V} = 0.
082 \, \text{atm K}^{-1} \] This implies \( V = 1 \, \text{L} \).
Now, we need to find the pressure P when the temperature \( T = 12.
2 \) K.
Using \( P = (\text{slope}) \times T \): \[ P = (0.
082 \, \text{atm K}^{-1}) \times (12.
2 \, \text{K}) \] \[ P = 0.
082 \times 12.
2 \, \text{atm} \] \( 0.
082 \times 12.
2 \approx 0.
082 \times (10+2.
2) = 0.
82 + 0.
082 \times 2.
2 \) \( 0.
082 \times 2 = 0.
164 \).
\( 0.
082 \times 0.
2 = 0.
0164 \).
\( 0.
164 + 0.
0164 = 0.
1804 \).
\( P \approx 0.
82 + 0.
1804 = 1.
0004 \, \text{atm} \).
So, \( P \approx 1.
0 \) atm.
Alternatively, using \( PV=nRT \) with \(V=1\) L, \(n=1\) mol, \(R=0.
082\), \(T=12.
2\) K: \( P \times (1 \, \text{L}) = (1 \, \text{mol}) \times (0.
082 \, \text{L atm mol}^{-1} \text{ K}^{-1}) \times (12.
2 \, \text{K}) \) \[ P = 0.
082 \times 12.
2 \, \text{atm} = 1.
0004 \, \text{atm} \approx 1.
0 \, \text{atm} \] This matches option (3).